FYI: students are permitted to use a graphing calculator on the test. Since the constants are real, one might set them all = 1 and graph. I would look like this, where the minimum value is -q since q = 1.

This is the minimum value or lowest point on a parabola not 0. the lowest point of this parabola is at k-q.
This is when the abs value is k and f(x) = k-q.
This is the vertex of the curve.
Go to KhanAcademy.org and watch the videos on Parabolas in the algebra 1 section. I have been doing this for the past 4 months. It is a great way to brush up on your math up thru calculus.

philomel: are you questioning if the answer is A which is -q? That's the correct answer. As you can see from the graph, the min value is -1. If q = 1, the min is -q.

Khan is a good source if you can stand to listen to him...I can't stand him, frankly. Also, his grammar leave much to be challenged and some of this math statements are actually INCORRECT! But, for most, he's a good source.

The vertical bars have always meant absolute value (well, since 1841, anyway).
That's not new notation!
The problem can be solve algebraically as has been stated or can be solved by graphing.
Students may use graphing calculators on the standardized test(s).

This is NOT advanced for 11th grade. Many students in high school today take precalculus as a sophomore and their first year of calculus as a junior.